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`int(cosx-sinx+1-x)/(e^(x)+sinx+x)dx=log_(e)(f(x))+g(x)+C` where C is the constant of integration and f(x) is positive. Then `f(x)+g(x)` has the value equal toA. `e^(x)+sinx+2x`B. `e^(x)+sinx`C. `e^(x)-sinx`D. `e^(x)+sinx+x` |
Answer» Correct Answer - B `I=int((e^(x)+cosx+1)-(e^(x)+sinx+x))/(e^(x)+sinx+x)` `=log_(e)(e^(x)+sinx+x)-x+C` `therefore" "f(x)=e^(x)+sin x+x and g(x)=-x` `" "f(x)+g(x)=e^(x)+sinx` |
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